Defining Random Variables and Probability Distributions - Statistics and Probability

A card is drawn at random from a set of 3 cards labeled A, B, and C. The sample space is $\{A,B,C\}$ with each outcome equally likely. Define the random variable $X$ as follows: $X(A)=0$ , $X(B)=0$ , and $X(C)=1$ . Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | $0$ | $\tfrac{2}{3}$ | | $1$ | $\tfrac{1}{3}$ | This problem involves a random variable where multiple outcomes map to the same value. A random variable X assigns numbers to outcomes in {A, B, C}, with X(A) = 0, X(B) = 0, and X(C) = 1. Since A and B both map to X = 0, we combine their probabilities: P(X = 0) = P(A) + P(B) = 1/3 + 1/3 = 2/3. Only C maps to X = 1, so P(X = 1) = 1/3. The correct distribution lists X-values 0 and 1 with probabilities 2/3 and 1/3, summing to 1. A common mistake is thinking each X-value must have equal probability, but this depends on the mapping. The approach is: identify all outcomes mapping to each X-value → sum their individual probabilities.
A bag contains 2 red marbles (R) and 1 blue marble (B). One marble is drawn at random. Define the random variable $X$ as the number of red marbles drawn. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | 0 | $\tfrac{1}{3}$ | | 1 | $\tfrac{2}{3}$ | The concept here is random variables and their probability distributions. A random variable assigns a numerical value to each possible outcome of a random experiment. In this marble drawing experiment, the outcomes are drawing a red or blue marble, and X maps red to 1 and blue to 0. The probability for each X value is the total probability of the outcomes mapping to it, so P(X=0) = P(blue) = 1/3, P(X=1) = P(red) = 2/3. The correct distribution in choice A lists all possible X values with probabilities summing to 1 and accurately reflecting the bag's composition. A common misconception is to confuse the distribution of outcomes (like colors) with the distribution of the numerical random variable X, as in choice B. To construct such distributions, list all possible outcomes with their probabilities, map each to its X value, and combine the probabilities for each distinct X.
From 3 red and 2 blue chips, draw 2 without replacement. X = number of red. Which distribution is correct?
Answer: $P_0=1/10,P_1=3/5,P_2=3/10$ There are 5 choose 2 = 10 equally likely pairs. Zero red means both blue: 2 choose 2 = 1 way, so 1/10. One red: 3 choices red times 2 blue = 6 ways, so 6/10 = 3/5. Two red: 3 choose 2 = 3 ways, so 3/10. The tempting wrong answer uses 4/25, 12/25, 9/25, which would be drawing with replacement.
A student flips two fair coins (coin 1 and coin 2). The sample space is $\{HH, HT, TH, TT\}$ . Define the random variable $X$ as the number of heads. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | $0$ | $\tfrac{1}{4}$ | | $1$ | $\tfrac{1}{2}$ | | $2$ | $\tfrac{1}{4}$ | This problem asks us to find the probability distribution for the number of heads when flipping two coins. A random variable X assigns numbers to outcomes, here counting heads in each outcome. The outcomes {HH, HT, TH, TT} map to X-values: X(HH) = 2, X(HT) = 1, X(TH) = 1, X(TT) = 0. Since HT and TH both give 1 head, we combine their probabilities: P(X = 1) = 1/4 + 1/4 = 1/2. The complete distribution is P(X = 0) = 1/4, P(X = 1) = 1/2, P(X = 2) = 1/4, which sums to 1. A common error is listing the outcomes (HH, HT, etc.) instead of the X-values (0, 1, 2). The strategy is: identify what X counts → evaluate X for each outcome → group by X-value and add probabilities.
A $3 ticket pays $10 on a roll of 6 and $0 otherwise. X is net gain. Which distribution is correct?
Answer: $P(X=7)=1/6, P(X=-3)=5/6$ A roll of 6 has probability 1/6; you win $10 but paid $3, so net gain is $7. Any other roll has probability 5/6 and pays $0, so net gain is -$3. The tempting mistake is using $10 as your gain, forgetting to subtract the ticket cost; $0 on losing rolls is also wrong because you still lose the $3.
A survey asks a randomly selected student how many siblings they have. Suppose the only possible responses in this group are 0, 1, or 2 siblings, with probabilities $P(0)=0.30$ , $P(1)=0.50$ , and $P(2)=0.20$ . Let $X$ be the number of siblings reported. Which graph represents the distribution of $X$ ?
Answer: A bar graph with bars at $X=0,1,2$ having heights $0.30, 0.50, 0.20$ respectively. Here, the concept is random variables and distributions represented graphically for sibling counts. A random variable assigns numbers to survey responses, with X as the number of siblings (0,1, or 2). Each response maps directly to X=0,1, or 2 with given probabilities 0.30, 0.50, 0.20. These probabilities are assigned to X values and visualized as bar heights. The graph in choice A is complete and valid, showing bars for all possible X (0,1,2) with correct heights summing to 1 implicitly via the probabilities. Misconception: confusing response categories with X values, but X is the numerical count. To create this, list possible responses, map to X, and plot combined probabilities as bar heights.
A jar contains 2 green marbles and 2 yellow marbles. You draw 2 marbles without replacement. Define the random variable $X$ as the number of green marbles drawn. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | 0 | $\tfrac{1}{6}$ | | 1 | $\tfrac{4}{6}$ | | 2 | $\tfrac{1}{6}$ | This question explores random variables and distributions for marble draws without replacement. A random variable assigns numbers to outcomes, with X as the count of green marbles in a draw of two from two green and two yellow. Possible outcomes are pairs like GG, GY, YY, mapping to X=2,1,0 respectively. Using combinations, P(X=0)=C(2,2)/C(4,2)=1/6, P(X=1)=[C(2,1)*C(2,1)]/6=4/6, P(X=2)=1/6. Choice A's distribution is complete and valid, covering X=0,1,2 with probabilities summing to 1 and reflecting the hypergeometric setup. Misconception: listing pairs like GY as X instead of the green count. Method: list all six pairs, map to X, combine probabilities.
A spinner has 4 equal sections labeled A, B, C, and D. You spin once. Define the random variable $X$ by: $X=2$ if the result is A or B, $X=5$ if the result is C, and $X=8$ if the result is D. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | 2 | $\tfrac{1}{2}$ | | 5 | $\tfrac{1}{4}$ | | 8 | $\tfrac{1}{4}$ | The key concept is defining random variables and distributions based on spinner outcomes mapped to custom values. A random variable assigns numerical values to outcomes, with X=2 for A or B, X=5 for C, and X=8 for D on a four-section spinner. Each spin outcome (A, B, C, D) maps to its corresponding X value, with equal probability 1/4 per section. Probabilities for X are aggregated: P(X=2)=P(A)+P(B)=1/2, P(X=5)=1/4, P(X=8)=1/4. The distribution in choice C is complete and valid because it covers all possible X (2,5,8), sums to 1, and correctly combines probabilities from multiple outcomes to each X. People often confuse outcomes like 'A' with X values, but X is the assigned number. To build this, list spinner outcomes, map to X, and sum probabilities for duplicate X values.
A fair six-sided die is rolled once. Define the random variable $X$ as the number of points you earn, where you earn 0 points if you roll 1 or 2, 1 point if you roll 3 or 4, and 2 points if you roll 5 or 6. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | 0 | $\tfrac{2}{6}$ | | 1 | $\tfrac{2}{6}$ | | 2 | $\tfrac{2}{6}$ | This example demonstrates random variables and distributions for points earned from a die roll. A random variable assigns numbers to outcomes, here X=0 for rolls 1-2, X=1 for 3-4, X=2 for 5-6. Each die face maps to an X value, with two faces per category. Since each face has probability 1/6, P(X=0)=2/6, P(X=1)=2/6, P(X=2)=2/6. The distribution in choice A is complete and valid as it includes all possible X (0,1,2), probabilities sum to 1, and it accurately groups the equal probabilities. A misconception is equating die faces directly to X without grouping. The approach: list all six outcomes, map to X, and combine probabilities for each X.
A jar contains 2 green balls and 2 yellow balls. Two balls are drawn without replacement. Let $X$ be the number of green balls drawn. Which table correctly represents the probability distribution of $X$ ?
Answer: $\begin{array}{c|c}X & P(X)\\\hline 0 & \tfrac{1}{6}\\ 1 & \tfrac{2}{3}\\ 2 & \tfrac{1}{6}\end{array}$ This problem involves drawing 2 balls without replacement from 2 green and 2 yellow balls, with X counting green balls drawn. A random variable X assigns numbers to outcomes - here it counts green balls in each draw combination. The possible outcomes and their X-values are: GG → X = 2, GY → X = 1, YG → X = 1, YY → X = 0. For probabilities without replacement: P(GG) = (2/4)(1/3) = 1/6, P(GY) = (2/4)(2/3) = 1/3, P(YG) = (2/4)(2/3) = 1/3, P(YY) = (2/4)(1/3) = 1/6. Combining same X-values: P(X = 0) = 1/6, P(X = 1) = 1/3 + 1/3 = 2/3, P(X = 2) = 1/6. Option A correctly shows this complete distribution. The key insight: without replacement changes probabilities for the second draw based on the first.
A fair spinner has 5 equal sections labeled A, B, C, D, and E. It is spun once. Define the random variable $X$ as the number of vowels spun, where vowels are A and E. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | $0$ | $\tfrac{3}{5}$ | | $1$ | $\tfrac{2}{5}$ | This problem asks us to count vowels in a spinner outcome. A random variable X counts the number of vowels spun, where A and E are vowels. The outcomes {A, B, C, D, E} map to X-values: X(A) = 1, X(B) = 0, X(C) = 0, X(D) = 0, X(E) = 1. Three outcomes (B, C, D) give 0 vowels, so P(X = 0) = 3/5. Two outcomes (A, E) give 1 vowel, so P(X = 1) = 2/5. The correct distribution lists X-values 0 and 1 with their probabilities, not the letters or categories. A common error is thinking about vowel/non-vowel categories instead of counting. The strategy is: identify what to count → evaluate the count for each outcome → combine probabilities for the same count.
A spinner has 4 equal sections labeled 1, 2, 3, and 4. The spinner is spun once. Define the random variable $X$ as the value shown. Which table correctly represents the probability distribution of $X$ ?
Answer: $\begin{array}{c|c}X & P(X)\\\hline 1 & \tfrac{1}{4}\\ 2 & \tfrac{1}{4}\\ 3 & \tfrac{1}{4}\\ 4 & \tfrac{1}{4}\end{array}$ This problem involves a spinner with 4 equal sections, where X represents the value shown. A random variable X assigns numerical values to experimental outcomes - here X equals the section number landed on. The spinner has sections labeled 1, 2, 3, and 4, so the possible X-values are {1, 2, 3, 4}. Since the sections are equal, each has probability 1/4: P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = 1/4. Option A correctly shows all four X-values with equal probabilities. Option D makes the common error of describing outcomes in words rather than showing X-values - probability distributions must show the numerical values of X, not descriptions. The approach is: identify possible X-values → determine their probabilities → verify probabilities sum to 1.
A spinner has 4 equal sections labeled 1, 2, 3, and 4. The spinner is spun once. Define the random variable $X$ as the amount of money won in dollars, where the winnings equal $2$ times the number landed on. Which description correctly defines the random variable $X$ ?
Answer: $X$ is the number of dollars won, taking values $\{2,4,6,8\}$ depending on whether the spinner lands on 1, 2, 3, or 4. The concept here is defining random variables and their probability distributions. A random variable assigns a numerical value to each possible outcome of a random experiment. In this spinner experiment, the outcomes are landing on 1,2,3, or 4, and X is defined as 2 times that number, so mapping to 2,4,6,8. The probabilities are assigned to these X values, each 1/4. The correct definition in choice B accurately describes X as the numerical winnings with values {2,4,6,8}. A common misconception is to think X is the labels or the probabilities themselves, as in choices A and D. To define such random variables, identify the outcomes, assign a numerical function to them, and then the distribution follows by mapping probabilities accordingly.
A bag contains 3 red marbles and 1 blue marble. One marble is drawn at random. Define the random variable $X$ as the number of red marbles drawn. Which table correctly represents the probability distribution of $X$ ?
Answer: $\begin{array}{c|c}X & P(X)\\\hline 0 & \tfrac{1}{4}\\ 1 & \tfrac{3}{4}\end{array}$ This problem asks us to define a probability distribution for a random variable X representing the number of red marbles drawn. A random variable assigns numerical values to the outcomes of a random experiment. Here, the outcomes are drawing either a red marble (R) or blue marble (B), and $X = 1$ if red is drawn, $X = 0$ if blue is drawn. Since there are 3 red marbles out of 4 total, $P(X = 1) = \frac{3}{4}$ and $P(X = 0) = \frac{1}{4}$ . The correct distribution lists all possible X-values with their probabilities, so option B is correct. A common mistake is confusing the outcomes (R, B) with the X-values (0, 1) - option C shows outcomes instead of X-values. To solve such problems: identify outcomes → map to X-values → calculate probabilities for each X-value.
A game uses one fair coin flip. If the result is Heads, you win $3; if the result is Tails, you lose $1. Let $X$ be your winnings (in dollars) from one play of the game. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | $-1$ | $\tfrac{1}{2}$ | | $3$ | $\tfrac{1}{2}$ | The concept here is random variables and their probability distributions. A random variable assigns a numerical value to each possible outcome of a random experiment. In this game, outcomes are Heads or Tails, mapped to X=3 or X=-1 respectively. Probabilities: P(X=3)=1/2, P(X=-1)=1/2. The correct distribution in choice A lists all possible X values with probabilities summing to 1 and matching the game rules. A common misconception is to list outcomes like Heads/Tails instead of numerical X, as in choice B. To construct such distributions, list all possible outcomes with their probabilities, map each to its X value, and combine the probabilities for each distinct X.
A survey randomly selects one student from a class. The student's grade level is equally likely to be 9th, 10th, 11th, or 12th. Define the random variable $X$ as the grade level number (9, 10, 11, or 12). Which table correctly represents the probability distribution of $X$ ?
Answer: $\begin{array}{c|c}X & P(X)\\\hline 9 & 0.25\\ 10 & 0.25\\ 11 & 0.25\\ 12 & 0.25\end{array}$ This problem involves selecting a student where X represents their grade level number. A random variable X assigns numerical values to outcomes - here X equals the grade level (9, 10, 11, or 12). Since each grade level is equally likely, the four outcomes have equal probability. With 4 possible outcomes, each has probability 1/4 = 0.25: P(X = 9) = P(X = 10) = P(X = 11) = P(X = 12) = 0.25. Option A correctly shows all four X-values with equal probabilities that sum to 1. Option D incorrectly uses ordinal labels (9th, 10th) instead of the numerical X-values - the random variable X is defined as the grade number itself. To verify: list all possible X-values → assign equal probabilities → ensure they sum to 1.
A fair six-sided die is rolled once. Define the random variable $X$ as the number of even results obtained in this one roll (so $X=1$ if the die shows an even number and $X=0$ otherwise). Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | 0 | $\frac{1}{2}$ | | 1 | $\frac{1}{2}$ | This problem defines X as the number of even results in one die roll, which can only be 0 or 1. A random variable assigns numbers to outcomes - here X indicates whether the die shows an even number. The die outcomes {1,2,3,4,5,6} map to X-values: X=0 for odd outcomes {1,3,5} and X=1 for even outcomes {2,4,6}. Since a fair die has equal probability 1/6 for each face, we calculate: P(X=0) = P(odd) = 3/6 = 1/2 and P(X=1) = P(even) = 3/6 = 1/2. Option A correctly shows this complete distribution with both possible X-values and their equal probabilities. The key insight is that in a single roll, you either get an even number (X=1) or you don't (X=0) - there's no possibility of getting multiple even numbers, so X cannot equal 2.
A student randomly selects one card from a set of 5 cards labeled A, B, C, D, and E (each card equally likely). Define the random variable $X$ as follows: $X=0$ if A or B is selected, $X=1$ if C is selected, and $X=2$ if D or E is selected. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | 0 | $\frac{2}{5}$ | | 1 | $\frac{1}{5}$ | | 2 | $\frac{2}{5}$ | This problem defines X based on which card is selected from five equally likely cards. A random variable assigns numbers to outcomes - here cards A,B,C,D,E are mapped to X-values. The mapping is: cards A and B both give X=0, card C gives X=1, and cards D and E both give X=2. Since each card has probability 1/5, we calculate: P(X=0) = P(A or B) = 1/5 + 1/5 = 2/5, P(X=1) = P(C) = 1/5, and P(X=2) = P(D or E) = 1/5 + 1/5 = 2/5. Option A correctly displays this complete distribution with all possible X-values and their probabilities summing to 1. The key is combining probabilities when multiple outcomes map to the same X-value. Option C incorrectly omits X=2, making it an incomplete distribution.
A prize wheel has 5 equal sections labeled $0, $0, $1, $2, and $5. You spin once. Let $X$ be the amount of money won. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | 0 | $\tfrac{2}{5}$ | | 1 | $\tfrac{1}{5}$ | | 2 | $\tfrac{1}{5}$ | | 5 | $\tfrac{1}{5}$ | The concept focuses on random variables and distributions for money won on a prize wheel. A random variable assigns numerical values to outcomes, with X as the dollar amount from sections labeled $0$ (twice), $1$ , $2$ , and $5$ . Each section maps to its X value, with probability $1/5$ per section. Probabilities combine for repeated values: $P(X=0) = 2/5$ , $P(X=1) = 1/5$ , $P(X=2) = 1/5$ , $P(X=5) = 1/5$ . Choice B's distribution is complete and valid, listing all distinct X (0,1,2,5), summing to 1, and properly aggregating the double $0$ probability. Misconception: treating labels like ' $0$ ' as non-numeric X, but X is the amount. Strategy: list five outcomes, map to X, combine probabilities for identical X.
A box contains 1 gold ticket and 2 silver tickets. One ticket is drawn at random. You win $10$ for a gold ticket and $4$ for a silver ticket. Let $X$ be the amount of money won. Which table correctly represents the probability distribution of $X$ ?
Answer: | $X$ | $P(X)$ | |---|---| | $10$ | $\tfrac{1}{3}$ | This example covers random variables and distributions for winnings from ticket draws. A random variable assigns numbers to outcomes, with $X=10$ for gold, $X=4$ for silver, from one gold and two silver tickets. Each ticket maps to its $X$ value, with $P$ (gold)=\frac{1}{3}$, $P $(silver)=\frac{2}{3}$ . Thus, $P(X=10)=\frac{1}{3}$ , $P(X=4)=\frac{2}{3}$ . Choice A's distribution is complete and valid, listing distinct $X$ ( $4$ , $10$ ) with probabilities summing to 1 and matching the draw odds. Misconception: using 'gold' or 'silver' as $X$ instead of monetary values. To derive: list three tickets, map to $X$ , combine probabilities for same $X$ (though none here).

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